3 improved formatting edit approved Aug 25 '16 at 19:05 Gilles 90411 gold badge88 silver badges2020 bronze badges The density for a normal distribution converges to 0 in the limit to both positive and negative infinity. So it is not true that $$ln(pdf(\theta_1)) > ln(pdf(\theta_0))$$$$\ln(\textrm{pdf}(\theta_1)) > \ln(\textrm{pdf}(\theta_0))$$ for all $$\theta_1$$ and $$\theta_0$$. The reason that the density is larger for this normal distribution near its mean compared to the uniform distribution is because the variance of the normal distribution is small. The density for a normal distribution converges to 0 in the limit to both positive and negative infinity. So it is not true that $$ln(pdf(\theta_1)) > ln(pdf(\theta_0))$$ for all $$\theta_1$$ and $$\theta_0$$. The reason that the density is larger for this normal distribution near its mean compared to the uniform distribution is because the variance of the normal distribution is small. The density for a normal distribution converges to 0 in the limit to both positive and negative infinity. So it is not true that $$\ln(\textrm{pdf}(\theta_1)) > \ln(\textrm{pdf}(\theta_0))$$ for all $$\theta_1$$ and $$\theta_0$$. The reason that the density is larger for this normal distribution near its mean compared to the uniform distribution is because the variance of the normal distribution is small. 2 added 1 character in body edited Jun 24 '15 at 13:54 TrynnaDoStat 5,67711 gold badge1515 silver badges3636 bronze badges The density for a normal distribution converges to 0 in the limit to both positive and negative infinity. So it is not true that $$ln(pdf(\theta_1)) > ln(pdf(\theta_0))$$ for all $$\theta_1$$ and $\theta_0$$\theta_0$$. The reason that the density is larger for this normal distribution near its mean compared to the uniform distribution is because the variance of the normal distribution is small. The density for a normal distribution converges to 0 in the limit to both positive and negative infinity. So it is not true that $$ln(pdf(\theta_1)) > ln(pdf(\theta_0))$$ for all $$\theta_1$$ and$\theta_0. The reason that the density is larger for this normal distribution near its mean compared to the uniform distribution is because the variance of the normal distribution is small. The density for a normal distribution converges to 0 in the limit to both positive and negative infinity. So it is not true that $$ln(pdf(\theta_1)) > ln(pdf(\theta_0))$$ for all $$\theta_1$$ and $$\theta_0$$. The reason that the density is larger for this normal distribution near its mean compared to the uniform distribution is because the variance of the normal distribution is small. 1 answered Jun 17 '15 at 16:42 jaradniemi 4,0461010 silver badges2424 bronze badges The density for a normal distribution converges to 0 in the limit to both positive and negative infinity. So it is not true that $$ln(pdf(\theta_1)) > ln(pdf(\theta_0))$$ for all $$\theta_1$$ and \$\theta_0. The reason that the density is larger for this normal distribution near its mean compared to the uniform distribution is because the variance of the normal distribution is small.